Difference between revisions of "Triangle inequality"

Revision as of 07:21, 27 April 2015 (view source) Alec (Talk | contribs) m ← Older edit		Revision as of 16:57, 28 August 2015 (view source) Alec (Talk | contribs) m (→‎Definition) Newer edit →
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	===Proof===		===Proof===
−	'''Style: ''' case analysis	+	We have 4 cases:
−		+	# Suppose that {{M|a>0}} and {{M|b>0}}
−	{{Todo|Take time to write it out}}	+	#: We see immediately that {{M|1=a>0\implies a+b>0+b=b>0}} so {{M|a+b>0}}
		+	#:* thus {{M|1=\vert a+b\vert=a+b}}
		+	#: We also see that {{M|1=\vert a\vert=a}} as {{M|a>0}}
		+	#: and that {{M|1=\vert b\vert=b}} for the same reason.
		+	#:* thus {{M|1=\vert a\vert+\vert b\vert=a+b}}
		+	#* We see that {{M|1=\vert a+b\vert=\vert a\vert+\vert b\vert}}
		+	#** Notice {{M|1=\vert a+b\vert=\vert a\vert+\vert b\vert\implies\vert a+b\vert\le\vert a\vert+\vert b\vert}} in the literal sense of "if left then right"
		+	#**: ({{M|\implies}} denotes [[Implies|logical implication]])
		+	# Suppose that {{M|a>0}} and {{M|b\le 0}}
		+	# Suppose that {{M|a\le 0}} and {{M|b> 0}}
		+	#* [[Mathematicians are lazy]], as {{M|\mathbb{R} }} is a [[field]] (an instance of a [[ring]]) we know that {{M|1=a+b=b+a}} (addition is [[commutative]])
		+	#* As {{M|\vert\cdot\vert:\mathbb{R}\rightarrow\mathbb{R} }} is a [[function]] we know that if {{M|1=x=y}} then {{M|1=\vert x\vert=\vert y\vert}}
		+	#* As, again, {{M|\mathbb{R} }} is a ring, we know that {{M|1=\vert a\vert+\vert b\vert=\vert b\vert+\vert a\vert}}
		+	#* So:
		+	#** {{M|1=\vert a+b\vert}}
		+	#**: {{M|1= =\vert b+a\vert}} by the [[Commutative|commutativity]] of addition on {{M|\mathbb{R} }}
		+	#**: {{M|1=\le \vert b\vert+\vert a\vert}} by the {{M|2^\text{nd} }} case (above)
		+	#**: {{M|1= =\vert a\vert+\vert b\vert}} again by commutativity of the real numbers
		+	#* Thus {{M|\vert a+b\vert\le \vert a\vert+\vert b\vert}}
		+	# Both {{M|a\le 0}} and {{M|b\le 0}}
		+	{{Todo|Finish proof}}
	==Reverse Triangle Inequality==		==Reverse Triangle Inequality==

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Revision as of 16:57, 28 August 2015

The triangle inequality takes a few common forms, for example:

$$d(x,z)\le d(x,y)+d(y,z)$$ (see metric space) of which $$|x-z|\le|x-y|+|y-z|$$ is a special case.

Another common way of writing it is

$$|a+b|\le |a|+|b|$$ , notice if we set $a=x-y$ and $b=y-z$ then we get $$|x-y+y-z|\le|x-y|+|y-z|$$ which is just $$|x-z|\le|x-y|+|y-z|$$

Definition

The triangle inequality is as follows:

• $$|a+b|\le |a|+|b|$$

Proof

We have 4 cases:

1. Suppose that $a>0$ and $b>0$

   We see immediately that $a>0\implies a+b>0+b=b>0$ so $a+b>0$

   • thus $\vert a+b\vert=a+b$

   We also see that $\vert a\vert=a$ as $a>0$

   and that $\vert b\vert=b$ for the same reason.

   • thus $\vert a\vert+\vert b\vert=a+b$

   • We see that $\vert a+b\vert=\vert a\vert+\vert b\vert$
     • Notice $\vert a+b\vert=\vert a\vert+\vert b\vert\implies\vert a+b\vert\le\vert a\vert+\vert b\vert$ in the literal sense of "if left then right"

       ($\implies$ denotes logical implication)

2. Suppose that $a>0$ and $b\le 0$
3. Suppose that $a\le 0$ and $b> 0$
   • Mathematicians are lazy, as $\mathbb{R}$ is a field (an instance of a ring) we know that $a+b=b+a$ (addition is commutative)
   • As $\vert\cdot\vert:\mathbb{R}\rightarrow\mathbb{R}$ is a function we know that if $x=y$ then $\vert x\vert=\vert y\vert$
   • As, again, $\mathbb{R}$ is a ring, we know that $\vert a\vert+\vert b\vert=\vert b\vert+\vert a\vert$
   • So:
     • $\vert a+b\vert$

       $=\vert b+a\vert$ by the commutativity of addition on $\mathbb{R}$

       $\le \vert b\vert+\vert a\vert$ by the $2^\text{nd}$ case (above)

       $=\vert a\vert+\vert b\vert$ again by commutativity of the real numbers

   • Thus $\vert a+b\vert\le \vert a\vert+\vert b\vert$
4. Both $a\le 0$ and $b\le 0$

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TODO: Finish proof

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Reverse Triangle Inequality

This is

$$|a|-|b|\le|a-b|$$

Proof

Take

$$|a|=|(a-b)+b|$$ then by the triangle inequality above:
$$|(a-b)+b|\le|a-b|+|b|$$ then $$|a|\le|a-b|+|b|$$ clearly $$|a|-|b|\le|a-b|$$ as promised

Note

However we see

$$|b|-|a|\le|b-a|$$ but $$|b-a|=|(-1)(a-b)|=|-1||a-b|=|a-b|$$ thus $$|b|-|a|\le|a-b|$$ also.

That is both:

• $$|a|-|b|\le|a-b|$$
• $$|b|-|a|\le|a-b|$$

Full form

There is a "full form" of the reverse triangle inequality, it combines the above and looks like:

$$|a-b|\ge|\ |a|-|b|\ |$$

It follows from the properties of absolute value, I don't like this form, I prefer just "swapping" the order of things in the abs value and applying the same result